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Answer:
the rate of heat loss is 2.037152 W
Explanation:
Given data
stainless steel K = 16 W 
diameter (d1) = 10 cm
so radius (r1) = 10 /2 = 5 cm = 5 × 
radius (r2) = 0.2 + 5 = 5.2 cm = 5.2 × 
temperature = 25°C
surface heat transfer coefficient = 6 6 W 
outside air temperature = 15°C
To find out
the rate of heat loss
Solution
we know current is pass in series from temperature = 25°C to 15°C
first pass through through resistance R1 i.e.
R1 = ( r2 - r1 ) / 4
× r1 × r2 × K
R1 = ( 5.2 - 5 )
/ 4
× 5 × 5.2 × 16 × 
R1 = 3.825 ×
same like we calculate for resistance R2 we know i.e.
R2 = 1 / ( h × area )
here area = 4
r2²
area = 4
(5.2 ×
)² = 0.033979
so R2 = 1 / ( h × area ) = 1 / ( 6 × 0.033979 )
R2 = 4.90499
now we calculate the heat flex rate by the initial and final temp and R1 and R2
i.e.
heat loss = T1 -T2 / R1 + R2
heat loss = 25 -15 / 3.825 ×
+ 4.90499
heat loss = 2.037152 W
Answer:
p=15.097lbf/in^2
Explanation:
the manometric pressure is that in which the atmospheric pressure is not taken into account, so for this case only the pressure exerted by the water on the bus is calculated using the following equation.
P=ρgh
where
ρ=density of water at 55°F=999.4kg/m^3
g=9.81m/s^2
h=35ft=10.668m
Solving
P=(994.4)(9.81)(10.668)=104067.02Pa = 15.097lbf/in^2
the gage pressure does a skin diver experience when they dive to 35 ft is 15.097lbf/in^2